A= (1+$\dfrac{a+\sqrt{a}}{1+\sqrt{a}}$)(1+$\dfrac{a-\sqrt{a}}{1-\sqrt{a}}$) với a>=0, a#1

Đáp án:

 `A=(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}).(1+\frac{a-\sqrt{a}}{1-\sqrt{a}})` `(đk: x\ge0;a\ne1)`

`=[1+\frac{\sqrt{a}.(1+\sqrt{a})}{1+\sqrt{a}}].[1-\frac{\sqrt{a}.(\sqrt{a}-1)}{\sqrt{a}-1}]`

`=(1+\sqrt{a}).(1-\sqrt{a})`

`=1^{2}-(\sqrt{a})^{2}`

`=1-a`

Vậy `A=1-a`